Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Note:

You may assume the interval's end point is always bigger than its start point.

Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

Example 1:

Input: [ [1,2], [2,3], [3,4], [1,3] ]

Output: 1

Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

Example 2:

Input: [ [1,2], [1,2], [1,2] ]

Output: 2

Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.

Example 3:

Input: [ [1,2], [2,3] ]

Output: 0

Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

 /**

  * Definition for an interval.

  * public class Interval {

  *     int start;

  *     int end;

  *     Interval() { start = 0; end = 0; }

  *     Interval(int s, int e) { start = s; end = e; }

  * }

  */

 public class Solution {

     public int eraseOverlapIntervals(Interval[] intervals) {

         if (intervals.length == 0) return 0;

         int nonOverlap = 1;

         int seq = 0;

         Arrays.sort(intervals, new Comparator<Interval>() {

             public int compare(Interval i1, Interval i2) {

                 return i1.end - i2.end;

             }

         });

         for (int i=1; i<intervals.length; i++) {

             if (intervals[i].start >= intervals[seq].end) {

                 seq = i;

                 nonOverlap++;

             }

         }

         return intervals.length - nonOverlap;

     }

 }

 Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[1], i2[1]));

 class Solution {

     public int eraseOverlapIntervals(int[][] intervals) {

         if (intervals.length < 1) return 0;

         int seq = 0;

         int nonOverlap = 1;

         Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[0], i2[0]));

         for (int i = 0; i < intervals.length; i ++) {

             if (intervals[i][0] >= intervals[seq][1]) {

                 seq = i;

                 nonOverlap ++;

             }

             else if (intervals[i][1] <= intervals[seq][1]) {

                 seq = i;

             }

         }

         return intervals.length - nonOverlap;

     }

 }